The answer is $k = 6$. First we show that $A$ cannot win for $k \ge 6$. Color the grid in three colors so that no two adjacent spaces have the same color, and arbitrarily pick one color $C$. $B$ will play by always removing a counter from a space colored $C$ that $A$ just played. If there is no such counter, $B$ plays arbitrarily. Because $A$ cannot cover two spaces colored $C$ simultaneously, it is possible for $B$ to play in this fashion. Now note that any line of six consecutive squares contains two spaces colored $C$. For $A$ to win he must cover both, but $B$'s strategy ensures at most one space colored $C$ will have a counter at any time.

Now we show that $A$ can obtain 5 counters in a row. Take a set of cells in the grid forming the shape shown below. We will have $A$ play counters only in this set of grid cells until this is no longer possible. Since $B$ only removes one counter for every two $A$ places, the number of counters in this set will increase each turn, so at some point it will be impossible for $A$ to play in this set anymore. At that point any two adjacent grid spaces in the set have at least one counter between them.

Consider only the top row of cells in the set, and take the lengths of each consecutive run of cells. If there are two adjacent runs that have a combined length of at least 4, then $A$ gets 5 counters in a row by filling the space in between. Otherwise, a bit of case analysis shows that there exists a run of 1 counter which is neither the first nor last run. This single counter has an empty space on either side of it on the first row. As a result, the four spaces of the second row touching these two empty spaces all must have counters. Then $A$ can play in the 5th cell on either side of these 4 to get 5 counters in a row. So in all cases $A$ can win with $k \le 5$.